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What is meant by the phrase "the mass is protected by a symmetry"?

+ 6 like - 0 dislike

In a particle physics context I've heard this phrase used. I guess it means that the mass of a particle is less than you'd naively expect from $E=mc^2$ after computing the momentum uncertainty corresponding to some appropriate length scale. Assuming my interpretation of the phrase is right, how does a symmetry let you get away with a smaller mass ?

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59
asked Apr 12, 2012 in Theoretical Physics by twistor59 (2,490 points) [ no revision ]
Many thanks for the answers everybody...

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59

4 Answers

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It means protection from radiative corrections. If you integrate loop corrections to the mass, out to some virtual momentum scale $\Lambda$, for a generic Lagrangian you would get contributions of order that scale. So what if you observe $m<<\Lambda$? Well if there is a symmetry in the theory than guarantees that perturbative corrections to the mass will vanish, then you say the mass is protected by the symmetry, problem solved. Otherwise you've got explaining to do why $m<<\Lambda$.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user user1631
answered Apr 12, 2012 by user1631 (60 points) [ no revision ]
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There are two ways in which this can happen. The trivial one is if the theory has some gauge invariance, in which case a mass term is simply not guage invariant and thus will never appear in any higher-order correction. There are complications if you break that symmetry and then combine it with some other field (e.g. electroweak or BCS transition).

The more complex possibility is via the Goldstone theorem. It states that if you break some continuous (global) symmetry then there will exist massless excitations. The broken symmetry version of that Lagrangian will often look very complex, and it will not be obvious that high-order diagrams will conspire to cancel and yield no radiative correction to the bare mass.

Concretely, consider a crystalline lattice. The lattice breaks translational invariance, and one obtains phonons as the Goldstone boson. In addition, these are guaranteed to have a dispersion relation which passes through zero. Intuitively, the reason is that one can apply a very long wavelength modulation to the atoms, and they will not exert a large potential on each other, so there will be very little restoring force.

A more interesting example is in QCD, where if the quarks were massless, then there would be an exact chiral symmetry. If that were so, then the $\pi$-mesons, which are the Goldstone bosons of breaking that symmetry, would be massless. However, the quarks are not quite massless (the bare quarks are on the order of a few tens of MeV) and thus the masslessness of $\pi$-mesons are not quite protected --- but they are still much less than the "natural" scale of confined QCD, which is on the order of a proton.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user genneth
answered Apr 12, 2012 by genneth (560 points) [ no revision ]
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In addition to the cases which genneth has mentioned,

  • Goldstone bosons: whose zero mass is protected by the broken symmetry
  • Gauge invariance: where charged chiral fermions can't have a mass by gauge invariance (charge conservation)

I have to add the following cases, which make the complete list:

  • Supersymmetry: (user1631's answer) This is the statement that a scalar particle is related by supersymmetry to a necessarily massless charged fermion. The fermion must be massless, the scalar (absent supersymmetry) doesn't have to be, but in order to have supersymmetry, the scalar ends up massless.
  • Chiral symmetry enforcement: This is the only new one here.

The last one is a little confusing, because it isn't the same as the chiral symmetry breaking by the QCD quark-condensate that makes the pions light. The chiral symmetry breaking by the QCD quark-condensate is just an ordinary Goldstone thing, and the pions are Goldstone bosons.

The chiral symmetry enforcement is the statement that I require that the Lagrangian of a fermion be invariant under rotations of the left and right chiralities in opposite directions. The mass term of a Dirac equation is not invariant under this, because the mass term produces a left chirality from a right chirality with a definite phase (you fix this phase by making m real and positive).

So in order to make a Dirac equation (with two chiralities) massless, you impose chiral symmetry, which prevents the two chiralities from turning into each other. This symmetry can be a gauged symmetry, in which case this is an example of Gauge invariance--- charge conservation doesn't allow the two to flip into each other--- but it can also be a non-gauged symmetry, and then you are just forcing the mass to zero to impose a symmetry.

This is confusing because there is a tendency in high energy physics to write all fermions in Dirac form, because people often memorize Dirac matrix formulas, not Weyl formulas, and Dirac formulas generalize to higher dimension easier. Then any 2-component chiral fermion is represented as two components of a Dirac spinor, but with an extra chiral symmetry which decouples the other 2 components. The other 2 components are not physical--- they aren't there--- but the form of the Lagrangian makes the massless condition come from the chiral symmetry.

This is the sense in which the term "mass protected by a symmetry" comes up the most--- when mass terms for a (global or gauge) charge carrying Fermion are disallowed because of their 2-component nature. It's not a good way to say it, it is best to only have 2 components in your head, and consider it forbidden just by charge conservation, not by a chiral symmetry, but people unfortunately use the symmetry language in this case all the time anyway.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Ron Maimon
answered Apr 13, 2012 by Ron Maimon (7,495 points) [ no revision ]
Having been interested (as I was several decades ago) more in the relativity side of things, spinors were 'always' two component $SL(2:\mathbb{C})$ objects. It took some adjusting the other way to get used to the Dirac versions!

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59
@twistor59: I agonized on this--- which is better notation? SL(2,C) is 4d specific, and in higher dimensions, you absolutely need Dirac, so Dirac is the most general. You need Majorana and Weyl conditions clearly separated in your head for 2d, 6d, 10d. But SL(2,C) is so much cleaner in 4d. I still don't have a good answer to the spinor notation problem--- you need both unfortunately. There will be a solution one day.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Ron Maimon
+ 0 like - 1 dislike

Nima gives a better explanation in this video I will be able to in written format here. Relevant portion starts at about 20 min in


along with some other interesting commentary.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user DJBunk
answered Apr 13, 2012 by DJBunk (80 points) [ no revision ]

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